public class Offer_12 {
    /*
    * 
    *
    请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。
    例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true
示例 2：

输入：board = [["a","b"],["c","d"]], word = "abcd"
输出：false
    * */
    public static boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if(dfs(board,i,j,words,0)){
                    return true;
                }
            }
        }
        return false;
    }

    private static boolean dfs(char[][] board, int i, int j, char[] words, int index) {
        if(i<0||i>=board.length||j<0||j>=board[0].length||words[index]!=board[i][j]){
            return false;
        }
        if(index==words.length-1){
            return true;
        }
        board[i][j]='.';
        char[][] newArray = copyArray(board);
        return dfs(newArray,i+1,j,words,index+1)||dfs(newArray,i-1,j,words,index+1)||dfs(newArray,i,j+1,words,index+1)||dfs(newArray,i,j-1,words,index+1);
    }

    private static char[][] copyArray(char[][] board) {
        char[][] newArray = new char[board.length][board[0].length];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                newArray[i][j] = board[i][j];
            }
        }
        return newArray;
    }

    public static void main(String[] args) {
//[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]]
        char[][] charArray = {{'A','B','C','E'},{'S','F','C','E'},{'A','D','E','E'}};
        String words = "ABCCED";
        System.out.println(exist(charArray,words));
    }
}
